The above results assume (lowest-order) single photon exchange is dominant. If two-photon exchan

The above results assume (lowest-order) single photon...

Key Concepts

Single Photon Exchange Approximation

This concept refers to the leading-order mechanism in electromagnetic scattering where the interaction is mediated by the exchange of a single virtual photon. In many scattering processes, this approximation simplifies calculations and predicts symmetric results for processes involving particles with the same type of charge magnitude but opposite sign, such as electrons and positrons, under the assumption that only this lowest order interaction is significant.

Two-Photon Exchange Processes

When higher-order effects are considered, contributions from diagrams involving the simultaneous exchange of two photons become relevant. These two-photon exchange processes introduce additional corrections that are not symmetric under charge conjugation, meaning that their contributions can differ in sign or magnitude when comparing particle versus antiparticle scatterings, thereby breaking the symmetry observed in the single photon exchange approximation.

Interference Effects in Scattering Amplitudes

Interference effects arise when amplitudes from different exchange processes, such as the single and two-photon exchanges, combine. The interference between these amplitudes can lead to modifications in the total cross section. Due to the difference in how these interference terms contribute for positively charged versus negatively charged particles, the net effect can cause the measurable cross sections for electron-proton and positron-proton scattering to differ, highlighting the importance of multi-photon processes in precise scattering experiments.

Transcript

00:01 In this particular question, we have to compare the wavelengths of an electron and a photon which are having considerable kinetic energy.

00:09 So we'll be using relativistic energy formula, relativistic energy for calculation of wavelength for the electron, right? and then we will be using similar concept for photon as well.

00:23 Now for the electron part, we have the relativistic energy given by e squared is equal to dc squared plus mc squared pole squared.

00:37 Now what is e? e is nothing but the kinetic energy of the electron, right? kinetic energy of electron.

00:49 So let me draw a schematic as well.

00:53 So let us assume the electron is traveling and it is having a energy ear.

00:59 Let's say, right? and what is p? p is the momentum of the electron, right? p is the momentum.

01:08 So let me write here p as well.

01:10 Now, what is c? c is the speed of light.

01:15 C is the speed of light.

01:21 And speed of light is a constant value.

01:23 So we will not, means play with numbers here.

01:26 We will be just playing with expressions to compare wavelengths of electron and photon.

01:31 So i'm just defining the terms first.

01:34 Now these are all the terms necessary and we have also m that is the mass of electron.

01:40 All right now after all these values we have another formula for calculating the wavelength and that formula is called d.

01:51 Brogley's formula right? so d brogley's formula tells you that wavelength of any particular particle is given by the ratio of planx constant a on the momentum of that particle, right? so here, h is nothing but blanks constant and p is the momentum as we have defined earlier.

02:15 All right.

02:16 Now, let us first find out p from this particular expression and then substitute it back to our lambda is equal to h by p formula.

02:26 All right? so first of all, from the above expression, we can find out the expression for p squared.

02:33 So p squared will be given by, here you can see first of all we have e squared and we will be subtracting m squared c to the power fourth right m c squared whole square i have opened that up and divided by the c squared that was multiplied with p squared right so that that eventually gets us to the point where we can find out p to be equivalent to under root e squared minus m squared c to the power fourth divided by c because in the denominator part we had c squared so upon taking the root will be only left with c right so let me tell it equation number two and the above one as equation number one now from one and two we have the formula for wavelength from one and two we can find out lambda that is a wavelength is equal to edge divided by p and what is p p is nothing but under root e squared minus m squared c to the power 4th and divided by c.

03:44 So i can just multiply it in the numerator part...

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